I miss to write conditions for one of the problems in the last comment. Regarding "Q4-(10)", a+b+c+d+e = a^3+b^3+c^3+d^3+e^3 = 5 (a, b, c, d, e: integers) (a≧b≧c≧d≧e) Find all (a, b, c, d, e). Find more than 6 solutions. (If there are less than 6 solutions, prove it.) is the correct problem. ...... The condition (a≧b≧c≧d≧e) was omitted. [ Excuse me. ]
hi someone can help me with this . just can’t seem to find the math method though i was able to brute force answer as 1232 Find the number of 2-element subsets {a, b} of {1,2,3,...,99, 100} such that ab + a + b is a multiple of 7.
One simplification. The required condition is equivalent to (ab + a + b + 1) is 1 mod 7, i.e., (a + 1) (b + 1) are inverses modulo 7. Call a + 1 = c, and b + 1 = d. Note that {c, d} is now a subset of {2, 3, ... , 101}. So now you're looking for the number of inverse pairs modulo 7 in the set {2, 3, ..., 101}. That should be quicker to count.
@@apratimroy hi - can you help me with this problem i am stuck with : prove 8x^3+12x^2+4x+1 is never a perfect square for natural numbers except for 2 and 1
@@infinity7858 , Hi, not sure, but this might be a helpful way to start. Suppose k^2 = 8 x^3 + 12 x^2 + 4 x + 1 = (2 x + 1)^3 - 2 x. Then k^2 - 1 = (2 x + 1)^3 - (2 x + 1). In other words, (k + 1) (k - 1) = u^3 - u = (u + 1) u (u - 1), where u = 2 x + 1 is an odd integer.
i dont get it why n >= 10, because [(4+1)/4]^3 is less then 2 or did i pluged it in the wrong equation. My english is bad so it can be that i didnt understand it correctly.
By induction... When n = 10, we have 2^10 = 1024, and 10^3 = 1000. Hence 2^n > n^3 for n = 10. Now, suppose we have 2^k > k^3 for some value k >= 10. Then 2^(k+1) > 2k^3. Since k >= 10, 1/k 2k^3 > (k+1)^3. Therefore, 2^(k+1) > (k+1)^3. Since it's true when n=10, by our inductive step, we know it's true for n=11. And since it's true for n=11, it is also true for n=12. Etc....
Thank you for explaining. Today's problem is an interesting integer problem. If you want to treat integer problems, I would like to inform as below: [information] I am also creating and uploading mathematics videos like you, so I would like to introduce you. In my case, with regard to mathematics, I had a time to study math for about only 1 to 2 years at university. When I entered university, I passed the faculty (that students can study normal math) of a national university, but my parents did not admit me to go to the university. It is a bitter experience. Therefore, after reaching retirement age, I started creating and uploading mathematics videos. However, I realized that there were some mistakes about using math words when I explain. (>_
I did another method and i found that (8^n+1)/(2^n+1)=4^n-2^n+1 And we know that 4^n-2^n+1 is an integer when n is a positive Thus (8^n+1)/(2^n+1) is an integer with all n such that n is a positive integer
oh i guess it must be if and only if, on the condition of (n^3-n )/(2^n+n) and the original question, so is final answer and doesn't need checking.
Very good. I solved it in much the same way.
It would be very easy to extend this to a solution for all integers, not just positive integers. n=0 is a solution. For n
prove is
The fraction becomes less than 2 at n = 8.
When n=8 it should be 504/264, but either way not an integer.
When n=8, the numerator should be 264 not 262
Genial.
Gràcies
I miss to write conditions for one of the problems in the last comment. Regarding "Q4-(10)",
a+b+c+d+e = a^3+b^3+c^3+d^3+e^3 = 5 (a, b, c, d, e: integers) (a≧b≧c≧d≧e)
Find all (a, b, c, d, e). Find more than 6 solutions. (If there are less than 6 solutions, prove it.)
is the correct problem. ...... The condition (a≧b≧c≧d≧e) was omitted. [ Excuse me. ]
i actually solved one for once!!
nice!
Can you solve: x^n + (2+x)^n + (2-x)^n = 0, x,n integers
Why is m=0 ruled out? It looks like (1+0)/(1+0)=1
0 is fine but probably not in natural numbers. (seems like that is the more standard interpretation of natural numbers, but I'm never quite sure.)
The answer is then the aliens
greets.😎
I wonder why need to subtract
Okay, I got it that must be the upper bound limit that must be divisible by the given divisor. 😅
hi someone can help
me with this . just can’t seem to find the math method though i was able to brute force answer as 1232
Find the number of 2-element subsets {a, b} of {1,2,3,...,99, 100} such that ab + a + b is a multiple of 7.
One simplification. The required condition is equivalent to (ab + a + b + 1) is 1 mod 7, i.e., (a + 1) (b + 1) are inverses modulo 7. Call a + 1 = c, and b + 1 = d. Note that {c, d} is now a subset of {2, 3, ... , 101}. So now you're looking for the number of inverse pairs modulo 7 in the set {2, 3, ..., 101}. That should be quicker to count.
@@apratimroy thank you so much . this was really helpful
@@infinity7858 , you're very welcome!
@@apratimroy hi - can you help me with this problem i am stuck with : prove 8x^3+12x^2+4x+1 is never a perfect square for natural numbers except for 2 and 1
@@infinity7858 , Hi, not sure, but this might be a helpful way to start. Suppose k^2 = 8 x^3 + 12 x^2 + 4 x + 1 = (2 x + 1)^3 - 2 x.
Then k^2 - 1 = (2 x + 1)^3 - (2 x + 1). In other words, (k + 1) (k - 1) = u^3 - u = (u + 1) u (u - 1), where u = 2 x + 1 is an odd integer.
Mathematical Prove Problem
i dont get it why n >= 10, because [(4+1)/4]^3 is less then 2 or did i pluged it in the wrong equation. My english is bad so it can be that i didnt understand it correctly.
By induction...
When n = 10, we have 2^10 = 1024, and 10^3 = 1000. Hence 2^n > n^3 for n = 10.
Now, suppose we have 2^k > k^3 for some value k >= 10. Then 2^(k+1) > 2k^3. Since k >= 10, 1/k 2k^3 > (k+1)^3. Therefore, 2^(k+1) > (k+1)^3.
Since it's true when n=10, by our inductive step, we know it's true for n=11. And since it's true for n=11, it is also true for n=12. Etc....
JAPAN MO 2009 P1
Thank you for explaining. Today's problem is an interesting integer problem. If you want to treat integer problems, I would like to inform as below:
[information]
I am also creating and uploading mathematics videos like you, so I would like to introduce you.
In my case, with regard to mathematics, I had a time to study math for about only 1 to 2 years at university.
When I entered university, I passed the faculty (that students can study normal math) of a national university,
but my parents did not admit me to go to the university.
It is a bitter experience. Therefore, after reaching retirement age, I started creating and uploading mathematics videos.
However, I realized that there were some mistakes about using math words when I explain. (>_
I did another method and i found that (8^n+1)/(2^n+1)=4^n-2^n+1
And we know that 4^n-2^n+1 is an integer when n is a positive
Thus (8^n+1)/(2^n+1) is an integer with all n such that n is a positive integer
You solved a different problem. This is (8^n + n) / (2^n + n) in Z, not (8^n + 1) / (2^n + 1) in Z.